Question

如果遇到android java编程的异步性问题，如何在onClick处理程序中抛出异常，或者如何使用警告对话框返回值？我必须使用变量来传递结果吗？

  public boolean validAccessCode = false;

//      public boolean requestAccessCode(Activity mActivity) throws Exception {
public void requestAccessCode(Activity mActivity) throws Exception {
  private EditText mPasswordView;

  mPasswordView = new EditText(mActivity);
  mPasswordView.setInputType(EditorInfo.TYPE_TEXT_VARIATION_PASSWORD);
  mPasswordView.setTransformationMethod(new PasswordTransformationMethod());

  AlertDialog.Builder alert = new AlertDialog.Builder(mActivity);
  alert.setTitle("Enter access code");
  alert.setView(mPasswordView);

  alert.setPositiveButton("Ok", new DialogInterface.OnClickListener() {
    public void onClick(DialogInterface dialog, int whichButton) {
      String pw = mPasswordView.getText().toString();
      if(pw.equals(accessCode))
      {
        validAccessCode = true;

         return;
      }

      //throw new Exception("Invalid access code");
    }

    alert.show();
  });

Answer 1

你可以在这里使用简单的界面

public interface DialogClickListener {
    public void onClicked(boolean validAccessCode)
}

public void requestAccessCode(Activity mActivity,
      final DialogClickListener  listener) throws Exception {
private EditText mPasswordView;

mPasswordView = new EditText(mActivity);
mPasswordView.setInputType(EditorInfo.TYPE_TEXT_VARIATION_PASSWORD);
mPasswordView.setTransformationMethod(new PasswordTransformationMethod());

AlertDialog.Builder alert = new AlertDialog.Builder(mActivity);
alert.setTitle("Enter access code");
alert.setView(mPasswordView);

alert.setPositiveButton("Ok", new DialogInterface.OnClickListener() {
public void onClick(DialogInterface dialog, int whichButton) {
  String pw = mPasswordView.getText().toString();

  if (listener != null) {
      listener.onClicked(pw.equals(accessCode));
  }
}

alert.show();

}）;

从带有警告对话框的方法返回值

1 个答案: